question bank maths class 10 circles

8/14/2019 question bank maths class 10 circles

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CIRCLES

A. SUMMATIVE ASSESSMENT

10.1 TANGENT TO A CIRCLE

1.A circle may be regarded as a collection of

points in a plane at a fixed distance from a fixed

point. The fixed point is called the centre of the

circle. The fixed distance between the centre of

the circle and the circumference, is calledradius.

2. The perimeter of the circle is referred to as

the circumference of the circle.

3.Achordof a circle is a line segment joining

any two points on the circumference.

4. An arc o f a c i r c l e i s a p a r t o f t h e

circumference.

5. A diameter of a circle is a chord which

passes through the centre of the circle.6. A line, which intersects the circle in two

distinct points, is called a secant.

7.A line which has only one point common to

the circle is called a tangent to the circle.

TEXTBOOKS EXERCISE 10.1

Q.1. How many tangents can a circle have?

Sol. A circle can have infinitely many tangents.

Since there are infinitely many points on a circle and at

each point of it, it has a unique tangent.

Q.2. Fill in the blanks :

(i) A tangent to a circle intersects it in

_______ point(s).

(ii) A line intersecting a circle in two points

is called a _______.

(iii) A circle can have _______ parallel

tangents at the most.

(iv) The common point of a tangent to a

circle and the circle is called _______.

Sol. (i) A tangent to a circle intersects it in one

point(s).

(ii) A line intersecting a circle in two points is

called a secant.

(iii) A circle can have twoparallel tangents at the

most.

(iv) The common point of a tangent to a circle

and the circle is called point of contact.

Q.3.A tangent PQ at a point P of a circle of

radius 5 cm meets a line through the centre O at

a point Q so that OQ = 12 cm. Length PQ is :

(a) 12 cm (b) 13 cm

(c) 8.5 cm (d) 119 cm

Sol. PQ is the tangent and OP is the radius.

OPQ = 90In right angled OPQ,

OQ2 = OP2 + PQ2 (By Pythagoras Theorem)

(12)2 = (5)2 + PQ2

144 = 25 + PQ2

PQ

2

= 144 25 PQ2 = 119

PQ 119 cm

Hence, the length PQ is 119 cm.

So, the correct choice is (d).

Q.4. Draw a circle and two lines parallel to

a given line such that one is a tangent and other,

a secant to the circle.

Sol. According to the given information we draw a

circle having O as centre andlis the given line.

10

Question Bank In Mathematics Class X (TermII)


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Now,mandnbe two lines parallel to a given line l

and that m is tangent as well as parallel to land n is

secant to the circle as well as parallel to l.

TEXTBOOKS EXERCISE 10.2

In Q.1 to 3 choose the correct option and give

justification.

Q.1. From a point Q, the length of the

tangent to a circle is 24 cm and the distance of

Q from the centre is 25 cm. The radius of thecircle is :

(a) 7 cm (b) 12 cm

(c) 15 cm (d) 24.5 cm

Sol.QPO = 90 [Angle between tangent andradius through the point of contact]

In right angledQPO, OQ2 = OP2 + PQ2

[By Pythagoras Theorem]

(25)2 = OP2 + (24)2

625 = OP2 + 576

OP2 = 625 576

OP2 = 49

OP 49 7 cm

So, radius of the circle is 7 cm.

Hence, the correct option is (a).

Q.2. In figure, if TP and TQ are the two

tangents to a circle with centre O so thatPOQ

= 110, then PTQ is equal to :

(a) 60 (b) 70

(c) 80 (d) 90

Sol.POQ = 110 (Given)

OPT = 90 andOQT = 90

[Angle between tangent and radius through

the point of contact is 90]

In quadrilateral OPTQ,POQ +OPT +OQT +PTQ = 360

[The sum of all the angles of a quadrilateral

is 360]

110 + 90 + 90 +PTQ = 360

290 +PTQ = 360

PTQ = 360 290 = 70

Hence, the correct choice is (b).

Q.3. If tangents PA and PB from a point P to

a circle with centre O are inclined to each other

10.2 NUMBER OF TANGENTS FROM A

POINT ON A CIRCLE

1. There is one and only one tangent at a

point of the circle.

2. The tangent at any point of a circle isperpendicular to the radius through the

point of contact.

3. No tangent can be drawn from a point

inside the circle.

4. The lengths of tangents drawn from an

external point to a circle are equal.

5. The perpendicular at the point of contact to

the tangent to a circle passes through thecentre of the circle.

6. Tangents drawn at the end points of a

diameter of a circle are parallel.


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at angle of 80, then POA is equal to :

(a) 50 (b) 60 (c) 70 (d) 80

Sol.OAP = 90 [Angle between tangent andradius through the point of contact]

1OPA BPA

2 [The centre lies on the bisector

of the angle between the two tangents]1

(80) 402

InOPA,OAP +OPA +POA = 180

[Sum of the three angles of a triangle is 180]

90 + 40 +POA = 180

130 + POA = 180

POA = 180 130

POA = 50

Hence, the correct choice is (a).

Q.4. Prove that the tangents drawn at the

ends of a diameter of a circle are parallel.

[2011 (T-II)]

Sol. PQ and RS are tangents to the circle C(O, r) atpoints A and B respectively, AB being diameter.

Since tangent is perpendicular to the radius at the

point of contact.

ABPQ and ABRS

PAB = 90 andABS = 90

PAB =ABS

PQ || RS [ PAB andABS are

alternate angles]Proved.

Q.5. Prove that the perpendicular at the

point of contact to the tangent to a circle passes

through the centre.

Sol. We know that the tangent at any point of acircle is perpendicular to the radius through the point of

contact and the radius essentially passes through the

centre of the circle, therefore the perpendicular at the

point of contact to the tangent to a circle passes through

the centre.

Q.6. The length of a tangent from a point A

at distance 5 cm from the centre of the circle is

4 cm. Find the radius of the circle. [Imp.]

Sol. We knowthat the tangent at any point of a circleis perpendicular to the radius through the point of

contact.

OPA = 90

In right angled

OPA, OA2

= OP2

+ AP2

[By Pythagoras Theorem]

(5)2 = OP2 + (4)2 25 = OP2 + 16

OP2 = 25 16 = 9 OP 9 3 cm

Hence, the radius of the circle is 3 cm.

Q.7.Two concentric circles are of radii 5 cm

and 3 cm. Find the length of the chord of the

larger circle which touches the smaller circle.

[Imp.]

Sol. Let O be the common centre of the twoconcentric circles.

Let AB be a chord of the larger circle which

touches the smaller circle at P.

Join OP and OA.

Then, OPA = 90

[ Th e t a n g e n t a t a n y p o i n t o f a c i rc l e i s

perpendicular to the radius through the point of contact]


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OBP = 90 ...(ii)

OAPB is a quadrilateral.

APB + AOB +OAP +OBP = 360

Putting the value ofOAP andOBP, we have

APB +AOB + 90 + 90 = 360

APB +AOB = 180

APB andAOB are supplementary.

Q.11.Prove that the parallelogram circums-

cribing a circle is a rhombus.

[2009, 2011 (T-II)]

Sol. Given : ABCD is a parallelogram circums-cribing circle.

To Prove : ABCD is a rhombus.Proof : Since the tangent segments from an


AP = AS ...(i)

BP = BQ ...(ii)

CR = CQ ...(iii)

and DR = DS ...(iv)

Adding equations (i), (ii), (iii) and (iv), we get

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AB + AB = AD + AD

2 AB = 2AD

AB = AD

But AB = CD and AD = BC

[Opposite sides of parallelogram] AB = BC = CD = AD

Hence, parallelogram ABCD is a rhombus.

Q.12. A t ri an gl e ABC i s d rawn t o

circumscribe a circle of radius 4 cm such that

the segments BD and DC into which BC is divided

by the point of contact D are of length 8 cm and

6cm respectively. (see figure). Find the sides AB

and AC. [HOTS]

Sol. Let the sides BC, CA and AB of the ABCtouch the circle at D, F and E respectively.

Join OE and OF. Also join OA, OB and OC.

Now, BD = BE = 8 cm and CF = CD = 6 cm

[ Tangent segments from an external point to a

circle are equal]

Also, let AE = AF =x cm. Then,

a= 14, b= x + 6, c= x + 8

Semi-perimeter ofABC2

a b c

14 ( 6) ( 8)( 14) cm

2

x xx

Area ofABC ( 14)( 14 14)x x

( 14 8)( 14 6) x x x x

( 14)( )(6)(8) x x

Now, area ofABC

= Area ofOBC + Area ofOCA

+ Area ofOAB

( 14)( )(6)(8)x x

(6 8)4 ( 6)4 ( 8)4

2 2 2

x x


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( 14)( )(6)(8) 28 2 12 2 16x x x x

( 14)( )(6)(8) 4 56x x x ( 14)( )(6)(8) 4( 14)x x x Squaring both sides, we have

(x+ 14)(x)(6)(8) = 16(x+ 14)2

3x= x + 14 3xx = 14

2x= 14 14

72

x

AB = x + 8 = 7 + 8 = 15 cm

and AC =x + 6 = 7 + 6 = 13 cm

Q.13. P ro ve th at o pp o site s ide s o f a

quadrilateral circumscribing a circle subtend

supplementary angles at the centre of the circle.[2011 (T-II)]

Sol. Given : ABCD is a quadrilateral circums-cribing a circle whose centre is O.

To Prove :(i)AOB +COD = 180

(ii)BOC +AOD = 180

Construction :Join OP, OQ, OR and OS.

Proof : Since tangent segments from an external

point to a circle are equal, therefore

AP = AS, BP = BQ, CQ = CR, DR = DS

InOPB andOBQ,

OP = OQ [Radii of the same circle]

OB = OB [Common]

BP = BQ [Proved above]

OBP OBQ [SSS congruence criterion]

1 =2 [CPCT]

Similarly, 3 =4,5 =6 and7 = 8

So, 1 +2 +3 +4 +5 +6 +7 +

8 = 360

1 +1 +4 +4 +5 +5 +8 +

8 = 360

2(1 +4 +5 +8) = 360

1 +4 +5 +8 = 180

(1 +8) + (4 +5) = 180

AOB +COD = 180

Similarly, we can prove that

BOC +AOD = 180. Proved.

OTHER IMPORTANT QUESTIONS

Q.1. The common point of the tangent and

the circle is called the :

(a) point of contact

(b) point of intersection

(c) point of concurrence

(d) none of the above

Sol.(a) The common point of the tangent and the

circle is called the point of contact.

Q.2. How many tangents of a circle passingthrough a point lying inside the circle?

(a) zero (b) one

(c) two (d) infinite

Sol. (a) There is no tangent to a circle passing

through a point lying inside the circle.

Q.3. How many tangents can we draw to a

circle from a point lying outside the circle?

(a) 0 (b) 1 (c) 2 (d) infinite

Sol. (a) There are exactly 2 tangents to a circle

through a point lying outside the circle.

Q.4. The tangents drawn at the ends of a

diameter of a circle are :

(a) parallel (b) perpendicular

(c) inclined at an angle of 60

(d) none of the above

Sol. (a) The tangents drawn at the ends of a

diameter of a circle are parallel.

Q.5. In the given figure, PT is a tangent to

the circle with centre O. If OT = 6 cm and


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OP = 10 cm, then the length of tangent PT is :

(a) 8 cm (b) 12 cm

(c) 10 cm (d) 16 cm

Sol. (a) By Pythagoras Theorem in OPT,

PT2 = OP2 OT2

PT2 = (10)2 (6)2 = 64 PT = 8 cm

Q.6. In a circle of radius 7 cm, tangent PTis drawn from a point P such that PT = 24 cm.

If O is the centre of the circle, then length of OP

is:

(a) 30 cm (b) 28 cm

(c) 25 cm (d) 31 cm

Sol.(c) OP2 = PT2 + OT2 = (24)2 + (7)2

= 576 + 49 = 625

OP 625 25 cm

Q.7.A point P is 26 cm away from the centre

of a circle and the length of tangent drawn from

P to the circle is 24 cm. The radius of the circleis :

(a) 8 cm (b) 10 cm

(c) 12 cm (d) 14 cm

Sol. (b) Let O be the centre of the circle and PT be

the tangent. Then, radius of the circle is OT.

OT2 = OP2 PT2 = (26)2 (24)2

= 676 576 = 100

OT2 = 100 OT 100 10 cm

Q.8. In the given figure, O is the centre of

two concentric circles of radii 3 cm and 5 cm.

AB is a chord of outer circle which touches theinner circle. The length of chord AB is :

(a) 4 cm (b) 8 cm

(c) 34 cm (d) 7 cm

Sol.(b) Draw OP AB. Join OA.Then, OP = 3 cm and OA = 5 cm.

OA2 = OP2 + AP2

AP2 = OA2 OP2 = 52 32

AP2 = 16 AP = 16 4 cm

AB = 2 AP = (2 4) cm = 8 cm

Q.9. In the figure, ABC is circumscribed

touching the circle at P, Q, R. If AP = 4 cm, BP

=6 cm, AC =12 cm and BC = x cm. Then, x = ?

(a) 10 cm (b) 6 cm

(c) 14 cm (d) 18 cm

Sol. (c) Since, the lengths of tangents drawn froma n e xt e rn al p oi nt t o a c ir c le a re e q ua l, t he n

AR = AP = 4 cm

CR = (AC AR) = (12 4) cm = 8 cm = CQ

And, BQ = BP = 6 cm

BC = BQ + CQ = (6 + 8) cm = 14 cm.

Q.10. In the given figure, quad. ABCD is

circumscribed, touching the circle at P, Q, R and

S. If AP =5 cm, BC =7 cm and CS = 3 cm, then

length AB = ?

(a) 9 cm (b) 10 cm

(c) 12 cm (d) 8 cm

Sol. (a) Since the lengths of tangents drawn froman external point to a circle are equal, then

AQ = AP = 5 cm

CR = CS = 3 cm, BR = (BC CR)


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= (7 3) cm = 4 cm

BQ = BR = 4 cm

AB = AQ + BQ = (5 + 4) cm = 9 cm.

Q.11. I n t he g iv en f i gu re , A BC D i s

circumscribed touching the circle at P, Q, R and

S. If AP = 6 cm, BP =5 cm, CQ = 3 cm and DR

= 4 cm, then perimeter of quad. ABCD is :

(a) 18 cm (b) 27 cm

(c) 36 cm (d) 22 cm

Sol. (c) Since the length of tangents from ane xt er na l p oi nt t o a c ir cl e a re e qu al , t he n

AP = AS = 6 cm, BP = BQ = 5 cm,

CR = CQ = 3 cm, DS = DR = 4 cm

Perimeter of quad. ABCD

= (AP + BP) + (BQ + CQ) + (CR + DR)

+ (AS + DS) = 36 cm.

Q.12. In the given figure, quad. ABCD is

circumscribed, touching the circle at P, Q, R and

S such thatDAB =90. If CS = 27 cm and CB

= 38 cm and the radius of the circle is 10 cm,

then AB = ?

(a) 17 cm (b) 28 cm(c) 19 cm (d) 21 cmSol.(d) Since the lengths of tangents drawn from

an external point to the circle are equal, thenCR = CS = 27 cm.So, BR = (BC CR) = (38 27) cm = 11 cm BQ = BR = 11 cm.Join OQ. Then, PAQO is a rectangle. AQ = PQ = 10 cm.Hence, AB = (AQ + BQ) = (10 + 11) cm = 21 cm.

Q.13. In the given figure, ABC is right-

angled at B such that BC = 6 cm and AB = 8 cm.

A circle with centre O has been inscribed inside

th e tr ia n gle . O P A B, O Q B C a nd

OR C. If OP = OQ = OR = x cm, then x = ?

[2011 (T-II)]

(a) 2 cm (b) 3 cm

(c) 2.5 cm (d) 4 cm

Sol.(a) AC2 = (AB2 + BC2) = 82 + 62

= 64 + 36 = 100

AC = 100 = 10 cmNow, CR = CQ = BC BQ = (6 x) cm

and AR = AP = (AB BP) = (8 x) cm

AC = CR + AR = (6 x) + (8 x)

= (14 2x)

14 2x= 10 x= 2 cm

Hence, radius of the circle = 2 cm

Q.14. Quadrilateral ABCD circumscribes a

circle as shown in figure. The side of the

quadrilateral which is equal to AP + BR is :

[2011 (T-II)]

(a) AD (b) AC

(c) AB (d) BC


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Sol.(c) Since the lengths of tangents drawn from

an external point to the circle are equal, then

AP = AQ and BR = QB.

AP + BR = AQ + QB = AB [AQ + QB = AB]

Q.15. In figure, if OC = 9 cm and OB = 15

cm, then BC + BD is equal to : [2011 (T-II)]

(a) 18 cm (b) 12 cm

(c) 24 cm (d) 36 cm

Sol.(c) InOCB, by Pythagoras theorem

CB = 2 2(15) (9) 225 81 144 = =

CB = 12 cm

BD = BC = 12 cm [length of tangents

drawn from an external point to the circle are equal]

Now, BC + BD = 12 cm + 12 cm

= 24 cm

Q.16. A tangent PA is drawn from an

external point P to a circle of radius 3 2 cmsuch that the distance of the point P from O is

6cm as shown in figure. The value ofAPO is:

[2011 (T-II)]

(a) 30 (b) 45

(c) 60 (d) 75

Sol.(b) LetAPO =.

In rightOAP, we have

sin=OA

OP

3 2 1sin

6 2 = =

45 =

Q.17. The figure, shows two concentric

circles with centre O. AB and APQ are tangents

to the inner circle from point A lying on the outer

circle. If AB = 7.5 cm, then AQ is equal to :

[2011 (T-II)]

(a) 18 cm (b) 15 cm

(c) 12 cm (d) 10 cm

Sol.(b) AP = AB = 7.5 cm [since the length of

tangents from an external

point to a circle are equal]

Now, AQ = 2AP = 2 7.5 cm = 15 cm

[since perpendicular from the centre bisects the chord]

Q.18. In figure, APB is a tangent to a circle

with centre O, at point P. If QPB = 50, then

the measure of POQ is : [2011 (T-II)]

(a) 120 (b) 100 (c) 140 (d) 150

Sol.(b) OPB = 90 [the tangent at any point

of a circle is perpendicular to the radius through the

point of contact]

OPQ = 90 QPB = 90 50 = 40

Now, since OP = OQ (radii)

OQP =OPQ = 40

40 + 40 +POQ = 180

POQ = 180 80 = 100

Q.19. In the given figure, the length of PR

is: [2011 (T-II)]

(a) 20 cm (b) 26 cm

(c) 24 cm (d) 28 cm


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Sol.(b) In rightOSR, OR2 = 52 + 122 = 169

OR = 169 OR = 13 cmIn rightPQO,

PO2 = 42 + 32 = 25PO= 25

PO= 5 cm

Now, OO= (3 + 5) cm = 8 cm

PR = PO + OO+ OR = 5 cm + 8 cm + 13 cm

= 26 cm

Q.20. CP and CQ are tangents to a circle

with centre O. ARB is another tangent touching

the circle at R. If CP = 11cm, BC = 7 cm, then

the length BR is : [2011 (T-II)]

(a) 11 cm (b) 7 cm

(c) 3 cm (d) 4 cm

Sol.(d) CP = CQ [length of tangents from

an exterior point to a circle are equal]

11 cm = QB + 7 cm

QB = 11 cm 7 cm = 4 cm BR = QB = 4 cm [length of tangents from

an exterior point to a circle are equal]

Q.21. In the given figure, O is the centre of

the circle. If PA and PB are tangents from an

external point P to the circle, then AQB is

equal to : [2011 (T-II)]

(a) 100 (b) 80

(c) 70 (d) 50

Sol.(d)PAO =PBO = 90

[Angle between tangent and

radius through the point of contact]

AOB = 360 (80 + 90 + 90)

= 360 260 = 100

Now,AQB =1

2AOB

[The angle subtended by an arc at the centre is

double the angle subtended by it at any point on the

remaining part o f the circle]

1100 50

2= =

Q.22. In the given figure, if AOB = 125,

then COD is equal to : [2011 (T-II)]

(a) 62.5 (b) 45

(c) 125 (d) 55

Sol. (d) If a c i rc l e t o u c h e s t h e si d e s o f a

quadrilateral, then the angles subtended at the centre by

a pair of opposite sides are supplementary. AOB +COD = 180

COD = 180 125 = 55

Q.23.In the given figure, AT is the tangent to the

circle with centre O such that OT = 4 cm and

OTA =30. Then AT is equal to : [2011 (T-II)]

(a) 4 cm (b) 2 cm

(c) 2 3 cm (d) 8 cm

Sol.(c) Let AT = x cm

cos 30 = 3

4 2 4 =

x x

x= 2 3 cm


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Q.24. In the given figure, CP and CQ are

tangents from an external point C to a circle with

centre O. AB is another tangent which touches

the circle at R. If CP = 11 cm and BR = 4 cm,

find the length of BC. [2010]

Sol.We have, CP = 11 cm, BR = 4 cm

Since, the tangent from an external point are equal,

we have CQ = CP = 11 cm and BQ = BR = 4 cm

BC = CQ BQ = (11 4) cm = 7 cm

Q.25. In the given figure, there are two

concentric circles with centre O and of radii 5

cm and3 cm. From an external point P, tangents

PA and PB are drawn to these circles. If AP =

12 cm, find the length of BP. [2010]

Sol.Join OA, OB and OP.

Then OA = 5 cm, OB = 3 cm and PA = 12 cm

InOPA, OP2 = OA2 + AP2

= 52 + 122

= 25 + 144 = 169

OP = 169 = 13 cm

Now, inOBP, OP2 = OB2 + BP2

BP2 = OP2 OB2 = (13)2 (3)2

= 160

= BP = 160 cm = 4 10 cm

Q.26. In the given figure, if ATO = 40,

findAOB. [2008]

Sol. InOAT andOBT,

4 =1 = 90 [Tangent isto the

radius through the point of contact]

OT = OT [Common]

OA = OB [Radius]

OAT OBT [RHS criterion] 3 =2 ....(i) [CPCT]

InOAT,3 +4 +5 = 180

[Angle sum property of a]

3+ 90 + 40 = 180

3 + 130 = 180 3 = 50 AOB =2 +3 =3 +3 [From (i)]

= 50 + 50 = 100.

Q.27. From a point P, the length of the

tangent to a circle is 15 cm and distance of P

from the centre of the circle is 17 cm. Then what

is the radius of the circle? [2008C]

Sol.OAP = 90

[Radius isto the tangent through the point of

contact]

In rightOAP, OA2 + AP2 = OP2

[Pythagoras Theorem]

r2 + (15)2 = (17)2 r2 + 225 = 289

r2 = 289 225 = 64r= 8 cm.

Q.28. The two tangents from an external

point P to a circle with centre O are PA and PB.


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If PB = 70, what is the value of OB?

[2008C]

Sol. 1 =2 = 90[Tangent isto the radius through the point of

contact]

In quad. OAPB,

AOB +1 +APB +2 = 360

[Angle sum property of a quad.] AOB + 90 + 70 + 90 = 360

AOB + 250 = 360

AOB = 360 250 = 110.

Q.29. In the figure, ABC is circumscribing

a circle. Find the length of BC. [2009]

Sol. We know that the lengths of tangents drawn

from an external point to a circle are equal.

So, AR = AQ = 4 cm ...(i)

BR = BP = 3 cm ...(ii)

CP = CQ = (AC AQ) = (11 4) cm

= 7 cm

So, BC = BP + CP = (3 + 7) cm = 10 cm.

Q.30. In the figure, CP and CQ are tangents

to a circle with centre O. ARB is another tangent

touching the circle at R. If CP = 11 cm and

BC = 7 cm, then find the length of BR. [2009]

Sol. We know that the lengths of tangents drawn

from an external point to a circle are equal.

So, CP = CQ ... (i)

AP = AR ... (ii)

BQ = BR ... (iii)

Now, CP = CQ

11= BC + BQ

11= 7 + BR [From (iii)]

BR= 11 7 = 4 cm.

Q.31. Two tangents PA and PB are drawn to

a circle with centre O from an external point P.

Prove thatAPB = 2OAB.

[2009, 2011 (T-II)]

Sol. OAP = 90 [Radius through the point

of contact is perpendicular to the tangent]

PAB = 90 OAB

InAPB, PA = PB [Tangents from an

external point to a circle are equal]

PBA =PAB = 90 OAB ...(i)

Also, APB = 180 (PBA +PAB)

= 180 2 (90 OAB) APB = 180 180 + 2OAB

APB = 2OAB Proved.

Q.32. ABC is an isosceles triangle, in which

AB = AC, circumscribed about a circle. Show

that BC is bisected at the point of contact.

[2008, 2011 (T-II)]

Sol. Given :An isosceles triangle ABC, in which

AB = AC, circumscribed about a circle.

To prove :BF = FC

Prove :We know that the tangents drawn from an



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So, AD = AE (i)

CE = CF (ii)

BD = BF (iii)

Also, AB = AC [Given]

AD+ BD = AE + CE

DB = CE (iv) [From (i)]

From (ii), (iii) and (iv), we have BF = FC

Hence, BC is bisected at the point of contact.

Proved.

Q.33. In the figure a circle is inscribed in a

quadrilateral ABCD in which B = 90. If AD =

23 cm, AB = 29 cm and DS = 5 cm, find the

radius (r) of the circle. [2008]

Sol. We know that the tangents drawn from an


AR = AQ (i)

DR = DS (ii)

BQ = BP (iii)AD = AR + DR

AD = AR + DS [From (ii)]

23 = AR + 5

AR = 23 5 = 18 cm

AB = AQ + BQ

AB = AR + BQ [From (i)]

29 = 18 + BQ

BQ = 29 18 = 11 cm (iv)

In quadrilateral OPBQ,

B = 90

OQB = OPB = 90

[Tangent is perpendicular to the radius at the point

of contact]

BQ = BP [From (iii)]

OQ = OP =r [Given]

OPBQ is a square.

OP = OQ = PB = BQ

r = 11 cm [From (iv)]

Q.34. In the figure, OP is equal to diameter

of the circle. Prove that ABP is an equilateral

triangle. [2008, 2011 (T-II)]

Sol. Suppose OP meets the circle at Q. Join QA

and AB.

We have, OP = diameter

OQ + PQ = diameter

PQ = diameter radius.

PQ = radius

So, OQ = PQ = radius

Thus, OP is the hypotenuse of right triangle OAP

and Q is the mid-point of OP.

OA = AQ = OQ

[ Mid-point of hypotenuse of a right triangle is

equidistant from the vertices]

OAQ is an equilateral triangle.

AOQ = 60

So, APO = 30

APB = 2APO = 60

Also, PA = PB PAB = PBA

But, APB = 60

Therefore, PAB = PBA= 60

Hence, APB is equilateral. Proved.

Q.35. In the figure, ADC = 90, BC =

38 cm, CD = 28 cm and BP = 25 cm. Find the

radius of the circle. [2006, 2011 (T-II)]

P

P


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Sol. Since tangent to a circle is perpendicular to

the radius through the point of contact.

OSD = ORD = 90, OR = OS

DROS is a square.

Als,o BP = BQ

[Tangents from an external point are equal]

BQ = 25 cm [BP = 25 cm]

BC CQ = 25

38 CQ = 25 [BC = 38 cm]

CQ = 38 25 = 13 cm

CR = CQ = 13 [CQ = 13 cm]

CD DR = 13 [CR = CD DR]

28 DR = 13 [CD = 28 cm]

DR = 28 13 = 15 cm

Since DROS is a square, so OR = DR = 15 cm.

Hence, radius of the circle = 15 cm.

Q.36. PQR is a right-angled triangle with

QR =12 cm and PQ = 5 cm. A circle with centre

O is inscribed in PQR. Find the radius of the

circle.

Sol.

Since OA PQ, OB QR and OA = OB

AOBQ is a square.

In right-angled triangle PQR,

PR2 = PQ2 + QR2

[Pythagoras theorem]

PR2 = 52 + 122

PR2 = 25 + 144

PR2 = 169

PR = 13 cm

PA = PC [Tangents drawn

from an external point are equal]

PQ AQ = PR RC

5 r= 13 RC

RC = 13 5 +r

RC = 8 +r

RB = RC [Tangents drawn

from an external point are equal]

QR BQ = RC 12 r= 8 + r

2r= 4 r= 2 cm

Hence, radius of the circle is 2 cm.

Q.37. A circle touches the side BC ofABC,

at P and touch AB and AC produced at Q and R

respectively. Prove that AQ =1

2Perimeter of

ABC. [2006, 2011 (T-II)]

Sol.Since tangents from an exterior point to acircle are equal in length.

BP = BQ ...(i)

CP = CR ...(ii)

and AQ = AR ...(iii)

Now, perimeter ofABC

= AB + BC + AC= AB + (BP + PC) + AC

= (AB + BQ) + (AC + CR)

[From (i) and (ii)]

= AQ + AR = 2AQ [AQ = AR]

So, AQ =1

2(Perimeter ofABC).

Q.38. In the figure, O is the centre of the

circle, PA and PB are tangent segments. Show

that the quadrilateral AOBP is cyclic.

[2011 (T-II)]Sol.OA AP and OB BP

[Tangent to a circle is perpendicular to the radius at

the point of contact.]


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OAP = 90 and OBP = 90

OAP + OBP = 90 + 90 = 180 ...(i)

OAPB is cyclic because in a quadrilateral if a

pair of opposite angles are supple mentary, then the

quadrilateral is cyclic. Proved.

Q.39. In the figure, circles C(O, r) and

C(O,2

r)touch internally at a point A and AB is

a chord of the circle C(O, r), intersect ing

C(O,2

r) at C. Prove that AC = CB. [HOTS]

Sol.Join OC, OA.

OCA is the angle in a semicircle.

OCA = 90 ... (i)

OC ABIn circle C(O,r), AB is chord and OC AB

[From (i)]

AC = CB

[Perpendicular from centre of circle to a chord

bisects the chord] Proved.

Q.40. If two tangents are drawn to a circle

from an external point, then (i) they subtends

equal angles at the centre, (ii) they are equally

inclined to the segment joining the centre to that

point. Prove. [HOTS]

Sol.

Given : PT and PT are two tangents drawn to a

circle C (O,r) from an external point P.

To prove : (i) POT = POTand

(ii) OPT = OPT

Proof :In POT and POT

PT = PT

[Tangents drawn from an external point]

OT = OT [Radii of the same circle]

PTO =PTO [Each = 90]

POT POT [SAS congruency]

POT = POT

and OPT = OPT [CPCT]

Q.41. In the figure below, the incircle of

ABC touches the sides BC, CA and AB at D, E

and F respectively. Show that :AF + BD + CE = AE + BF + CD

=1

2(Perimeter of ABC)

[Imp.]

Sol. We know that the tangents from an exterior

point to a circle are equal in length.

AF = AE ...(i)

BD = BF ...(ii)

CE = CD ...(iii)

Adding (i), (ii) and (iii), we get

AF + BD + CE = AE + BF + CD

Perimeter ofABC = AB + BC + AC

= (AF + FB) + (BD + CD) + (AE + EC)= (AF + AE) + (BF + BD) + (CD + CE)

= (2AF + 2BD + 2CE)

= 2(AF + BD + CE) [From (i), (ii) and (iii)]

AF + BD + CE =1

2 (Perimeter ofABC)

Hence, AF + BD + CE = AE + BF + CD

=1

2 (Perimeter ofABC) Proved.


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Q.42. In the figure, if AB = AC, prove that

BE = EC. [2011 (T-II)]

Sol. We know that the tangents from an exterior

point to a circle are equal in length.

AD = AF ...(i)

BD = BE ...(ii)

CE = CF ...(iii)

Now, AB = AC

AB AD = AC AD

[Subtracting AD from both sides]

AB AD = AC AF [Using (i)]

BD = CF

BE = CE Proved.

[Using (ii), (iii)]

Q.43. PQ is a chord of length 8 cm of acircle of radius 5 cm. The tangents at P and Q

intersect at a point T. Find the length TP.

[HOTS]

Sol.Let TR = y.

Since OT is perpendicular bisector of PQ.

PR = QR = 4 cm

In right triangle ORP, we have

OP2 = OR2 + PR2

OR2 = OP2 PR2 = 52 42 = 9

OR = 3 cm

In right triangles PRT and OPT, we have

TP2

= TR2

+ PR2

...(i)and, OT2 = TP2 + OP2

OT2 = (TR2 + PR2) + OP2 [From (i)]

(y+ 3)2 =y2 + 16 + 25

6y = 32

y =16

3

TR =16

3

From (i), we have

TP2 =

2216 256 4004 16

3 9 9

TP = 20

3cm

Q.44. Let A be one point of intersection of

two intersecting circles with centres O and Q.

The tangents at A to the two circles meet the

circles again at B and C, respectively. Let the

po int P be loc at ed so tha t AO PQ is a

parallelogram. Prove that P is the circumcentreof the triangle ABC. [HOTS]

Sol. If P is the circumcentre ofABC, then OP and

PQ are perpendicular bisectors of AB and AC

respectively. i.e., we have to show that OP and PQ are

perpendicular bisectors of AB and AC respectively.

We have, OA AC

[ Tangent to a circle isto the radius through

the point of contact]

PQ AC

[ OAQP is a parallelogram OA || PQ]

PQ is the perpendicular bisector of AC

[ Q is the centre of the circle]


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Similarly,

BA AQ

BA OPAQPO is a parallelogram

OP || AQ

OP is the perpendicular bisector of AB.

Thus, P is the point of intersection of perpendicular

bisectors PQ and PO of sides AC and AB respectively.

Hence, P is the circumcentre ofABC.Proved.

Q.45. If AB, AC, PQ are tangents in figure

below and AB = 5 cm, find the perimeter of

APQ. [2011 (T-II)]

Sol.We have, PB = PX, QC = QX and AB = AC

Perimeter ofAPQ

= AP + PQ + AQ

= AP + (PX + XQ) + AQ

= (AP + PX) + (AQ + XQ)

= (AP + PB) + (AQ + QC)

= AB + AC = (5 + 5) cm = 10 cm

Q.46. In figure below, PQ is tangent at a

point R of the circle with centre O.

IfTRQ = 30, findPRS. [Imp.]

Sol.We have,TRQ = 30

Since ST is a diameter and angle in a semicircle

is a right angle.

SRT = 90

Now,TRQ +SRT +PRS = 180

[Straight angle]

30 + 90 +PRS = 180

PRS = 180 120 = 60.

Q.47.In figure below, a circle touches all the

four sides of a quadrilateral ABCD with AB = 6

cm, BC = 7 cm and CD = 4 cm. Find AD.

[2011 (T-II)]

Sol. Since tangents drawn from an exterior point to

the circle are equal.

AP = AS ...(1)

BP = BQ ...(2)

CR = CQ ...(3)

and DR = DS ...(4)

Adding (1), (2), (3) and (4), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR)

= (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

6 + 4 = AD + 7 AD = 3 cm.

Q.48. In the figure, two circles touch each

other externally at C. Prove that the common

tangent at C bisects the other two common

tangents. [2007, 2011 (T-II)]


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Sol. Given : Two circles with centres at A and B

touch each other externally at C. RS and PQ are two

common tangents to the circles.

To prove : (i) PE = EQ

(ii) RF = FS.

Proof :We know that the tangents drawn from an

exterior point are equal.

PE = EC (1)

Similarly, EQ = EC (2)

PE = EQ [From (1) and (2)]

EC bisects PQ at E.

Similarly, we can prove that F is the bisector of RS.

So, EF bisects PQ and RS. Proved.

Q.49. Prove that the lengths of tangents

drawn from an external point to a circle are

equal. Using the above, prove the following :

q ua dr il at er al A BC D i s d ra wn t o

circumscribe a circle (see fig.). Prove that

AB + CD = AD + BC. [2008]

Sol.Part I

Given : PT and PT are two tangents drawn from an

external point P to a circle C (O,r).

To prove :PT = PT

Construction :Join TO, TO and PO.

Proof :InPOT andPOT

PO =PO (Common)

OT = OT (Radii of the same circle)

PTO =PTO = 90 [OT PT and OT PT']

POT POT (RHS congruence)

PT = PT (CPCT)Proved.

Part II :See Q.8(Textbooks Exercise10.2)

Q.50. Prove that the lengths of the tangents

drawn from an external point to a circle are

equal.

Using the above, do the following :

In the figure, TP and TQ are tangents from T

to the circle with centre O and R is any point on

the circle. If AB is a tangent to the circle at R,

prove that

TA + AR = TB + BR. [2008, 2011 (T-II)]

Sol. Part I : Same as Q.49.

Part II

From part I, we have

TP = TQ (i)

AP = AR (ii)

BQ = BR (iii)

From (i), TA + AP = TB + BQ

TA + AR = TB + BR [From (ii) and (iii)]

Proved.

Q.51. Prove that the tangent at any point of

a circle is perpendicular to the radius through

the point of contact.

Using the above, do the following :

In the figure, O is the centre of the two

concentric circles. AB is a chord of the larger

circle touching the smaller circle at C. Prove

that AC = BC. [2009, 2011 (T-II)]


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1. A point P is 13 cm from the centre of a

circle. Radius of the circle is 5 cm. Then,

the length of the tangent drawn from P to the

circle is :

(a) 10 cm (b) 11 cm(c) 12 cm (d) 13 cm

2. The length of the tangent drawn from a

point, whose distance from the centre of a circle

is 20 cm and radius of the circle is 16 cm is :

(a) 12 cm (b) 144 cm

(c) 169 cm (d) 25 cm

3. A tangent PQ at a point P of a circle of

radius 15 cm meets a line through the centre O

at a point Q so that OQ = 25 cm. Length of PQ

is:

(a) 5 cm (b) 25 cm

(c) 16 cm (d) 20 cm4. In the figure, if AD, AE and BC are

tangents to the circle at D, E and F respectively,

then :

(a) AD = AB + BC + CA

(b) 2AD = AB + BC + CA

(c)AD

2 = AB + BC + CA

(d) AD

4= AB + BC + CA

5. Two circles touch each other externally at

C and AB is a common tangent to the circles.

ACB is :

(a) 60 (b) 45 (c) 90 (d) 180

6. PQ is a tangent drawn from a point P to

a circle with centre O and QOR is a diameter of

the circle such thatPOR = 120.OPQ is :

(a) 50 (b) 40 (c) 30 (d) 25

7. In the figure, a circle touches all the foursides of a quadrilateral ABCD, whose three sides

are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find

AD. [2011 (T-II)]

PRACTICE EXERCISE 10.1A

Sol.Part I :

Given : A circle (O,r) and a tangent XY at P.

To prove :OP XY

Construction :Take any point Q on XY. Join OQ

which intersects the circle at R.

Proof : OP = OR (Radii of the circle)

Now, OQ = OR + RQ

OQ > OR

OQ > OP (OP = OR)

OP < OQ

Thus, OP is shorter than any other point on XY.

Hence, OPXY

(Perpendicular distance is the shortest distance

from a point to a line) Proved.

Part II :

Join OC. Then,

OCB = 90 [From part I]

For outer circle, AB is a chord

AC = BC

[Perpendicular from the centre bisects the chord]

Proved.


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17.In the figure, O is the centre of the circle,

AB is a diameter and AD is a tangent. IfBOC

= 120, findADC.

18. In the figure, PA and PB are the tangentsegments to a circle with centre O. Show that A,

O, B, P are concyclic.

19. Prove that in two concentric circles, the

chord of the larger circle which touches the

smaller circle is bisected at the point of contact.20. Prove that the line segment joining the

point of contact of two parallel tangents to a

circle is a diameter of the circle. [2011 (T-II)]

21. Prove that there is one and only one

tangent at any point on the circumference of a

circle.

22. From a point P, two tangents PA and PB

are drawn to a circle C(O, r). If OP = 2r, show

thatAPB is equilateral. [Imp.]

23. In the given figure PO QO. The

tangents to the circle with centre O at P and Q

intersect at a point T. Prove that PQ and OT are

right bisectors of each other. [2011 (T-II)]

24. If from an external point B of a circle

with centre O, two tangents BC and BD are

drawn such that DBC = 120, prove thatBO = 2BC. [2011 (T-II)]

B. FORMATIVE ASSESSMENT

Activity

Objective : To show the following with the help of an activity.

(a) Lengths to tangents drawn from an external point are equal.

(b) Tangents are equally inclined to the segment joining the centre to that point.

Materials Required : White sheet of paper (or rice paper), Geometry box, Sketch pens, A pair

of scissors.


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Figure 2

Figure 1

Figure 3

Preparation for the Activity

1. Draw a circle of any radius with centre 'O' on a rice paper.2. Take a point, say P, outside of the circle.

3. From point P, draw a line segment touching the circle at A

( th e p oi nt o f c on ta ct ), w hi ch i s t he r eq ui re d t an gen t.

[See figure 1]

4. Following step 3, draw one more tangent PB, B being the

point of contact. [See figure 1]

5. Join OA, OP and OB. [See figure 1]

(a) Lengths of tangents drawn from an external point are equal.

Presentation : Fold the paper along OP.

Observation :

You will observe that point A coincides with point B and

line segment PA (the tangent from point P) concides with line

segment PB (another tangent from the same point on the circle).

[See figure 2]

Thus, length of PA = length of PB.

Hence, lengths of tangents drawn from an external point are equal.

The above property can be verified a number of times by taking external point at differentlocations.

Result : Lengths of tangents drawn from an external point are equal.

(b) Tangents are equally inclined to the line segment joining the centre with the external

point.

1. Cut out the two triangles, OPA andOPB, so formed figure 3.

2. Colour the two triangles with different colours.

3. Put one triangle on the other.

Observation :

You will observe that two triangles are congruent to each other

(i.e., one triangle exactly superimposes the other) with the

following (angle) correspondence.

OPA =OPB, (Proves the required result)

PAO =PBO,

AOP =BOP (See figure 4)


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Figure 4

Hence, tangents drawn from an external point, equally

inclined to the line segment joining the centre with that point.Result : Tangents drawn equally inclined to the line segment

joining the centre with the external point.

Note :The tangent at any point of a circle is perpendicular to

the radius through the point of contactOAP =OBP = 90.

ANSWERS

A. SUMMATIVE ASSESSMENTPractice Exercise 10.1A

1. (c) 2. (a) 3. (d) 4. (b) 5. (c) 6. (c)

7. 3 cm 8. 28 cm 9. 10.9 cm 10. 10 cm 11. 24 cm 12. 5 cm

13. 7 cm, 5 cm, 3 cm 14. 8 2cm 15. 140 16. 8 cm 17. 60


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question bank maths class 10 circles

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